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1/21/2014 1 Intermolecular forces, Liquids and Solids Ch. 10.

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1 1/21/ Intermolecular forces, Liquids and Solids Ch. 10

2 1/21/20142 Some Characteristics of Gases, Liquids and Solids and the Microscopic Explanation for the Behavior GASLIQUIDSOLID Assumes shape/ volume of container –particles move past one another Assumes shape of container –particles move/ slide past one another Retains fixed volume /shape –rigid particles locked in place Compressible-lots of free space between particles Not easily compressible - little free space between particles Flows easily-particles move past one another Flows easily-particles move/ slide past one another Does not flow easily-rigid particles cannot move/slide past one another Diffusion within gas occurs rapidly-expand easily to fill available space Diffusion within liquid occurs slowly-does not expand to fill container Diffusion within solid occurs extremely slowly Not denseMore dense than gasesMore dense than liquid No significant attractive forces between molecules Significant attractive forces between molecules Most significant forces between molecules

3 1/21/20143 Intermolecular Forces Ideal gas law used to describe gases – –Volume of gas molecule is too small 99.9% of gas is empty space – –Gas molecules so far apart No significant intermolecular attraction – –If gases truly ideal (zero volume/ attractive forces), couldnt condense them to liquids

4 1/21/20144 To condense real gases – –Intermolecular attractive forces must overcome KE of gas molecules Increasing attractive forces by – –Decreasing distance between molecules – –Increasing pressure which forces gas molecules closer together Decreasing temperature lowers average KE

5 1/21/ Three recognized types of intermolecular attractions Intramolecular Intramolecular

6 1/21/20146 Ô Intermolecular forces Ô Intermolecular forces (all electrostatic) Ô van der Waals forces Ô van der Waals forces (developed equation for predicting deviation of gases from ideal behavior) Ô Dipole-Dipole attraction Ô Dipole-Dipole attraction: dipole molecules orient themselves so that +/ ends close to each other Ô London Dispersion Forces Ô Ô Relatively weak forces that exist among noble gas atoms/ nonpolar molecules. (Ar, C 8 H 18 ) Ô Ô Caused by instantaneous dipole, in which electron distribution becomes asymmetrical Ô Ô Ease with which electron cloud of atom can be distorted- polarizability Ô Hydrogen bonds F/O/N Ô Hydrogen bonds: dipole-dipole attraction in which H is bound to highly electronegative atom (F/O/N) Ô Ion-dipole force Ô Ion-dipole force (solutions-also electrostatic)

7 Ion-Dipole Forces Exists between ion and partial charge on end of polar molecule Important for solutions of ionic substance in polar liquids –+ ions surrounded by - ends of water molecules –- ions surrounded by + ends of water molecule 1/21/20147

8 Dipole-Dipole Forces Dipole-Dipole Forces (about 1% strength of covalent bonds) Weaker than ion-dipole forces dipolesNeutral polar molecules (dipoles) attract each other with δ+ / δ- ends Responsible for mutual solubility of polar molecules, such as NH 3 in H 2 O Explains how and why polar molecules may be condensed to liquid state 1/21/20148

9 9 London Forces (dispersion forces, instantaneous dipole forces, induced dipole forces) Weakest forces of attraction Main form of attraction in all nonpolar molecules – –No other type of van der Waals forces exist Nonpolar He/Ar can be condensed, so some kind of attractive interactions exist instantaneousmomentaryMovement of electrons within electron cloud cause instantaneous or momentary dipole moment (self-polarization) – –Resultant dipole induces polarization in neighboring molecules – –Mutual attraction of opposite ends of neighboring dipoles

10 1/21/201410

11 1/21/ As long as molecules are close together, this electron movement can occur over huge numbers of molecules Whole lattice of molecules can be held together in solid using van der Waals dispersion forces An instant later, distribution of electrons shifts

12 1/21/ Hydrogen Bonding Hydrogen Bonding special dipole-dipole attraction Hydrogen covalently bonded to highly electronegative elements (N, O, F) Bond strength higher than other dipole-dipole attractions Important in bonding of molecules (water/DNA)

13 1/21/201413

14 1/21/ Strength of dispersion forces Much weaker than covalent bonds – –Size of attraction varies considerably with size/shape of molecule In comparing intermolecular forces, molar mass is 1 st consideration – –When molar masses comparable, differences in intermolecular attractions mainly due to differences in molecular polarity (strengths of dipole attraction) – –When molar masses differ greatly, differences in intermolecular attractions mainly due to differences in molar masses themselves (strength of London forces)

15 1/21/ – –Magnitude of self-polarization increases with increasing numbers of electrons Polarizability indicates ease electron cloud of atom can be distorted to give dipolar charge distribution – –Large atoms with many electrons (electrons loosely held) exhibit higher polarizability than small atoms (electrons tightly held near nucleus) As atomic # increases, # electrons increases – –More electrons, more distance they can move – –Increased chance of temporary dipole and dipole interactions – –Bigger the dispersion forces

16 1/21/ Gaseous polar molecules show little attraction for each other (far apart) – –Rapidly become weaker as distance between dipoles increases – –Unimportant in gas phase due to distance between molecules In liquids and solids, molecules10x closer – –Boiling point (condensation point) indicative of attractive forces between molecules Measure of how much KE must be increased so that it can overcome attractive forces in liquid Greater attractive forces/polarity, higher BP Low boiling point indicates low attractive forces

17 1/21/ – –BP/MP increase going down group # es/atomic radius increase – –Why I is solid/Br liquid at room temperature – –Why all halogens have much higher BPs than their neighboring noble gases More electrons = more distance they move = bigger temporary dipoles (stickier) = bigger dispersion forces – –Because of greater temporary dipoles, xenon molecules are "stickier" than smaller neon molecules – –Neon molecules will break away from each other at much lower temperatures than xenon molecules – neon has lower boiling point

18 1/21/ Shape of molecules – –More linear molecules w/greater contact between molecules have higher BP than more spherically shaped molecules Develop bigger temporary dipoles due to electron movement than short fat ones containing same numbers of electrons Long thin molecules lie closer together-attractions most effective if molecules really close – –Responsible for mutual solubility of nonpolar molecules such as Br 2 in CCl 4 Large electron clouds easily polarized, so they have higher BPs than neighboring noble gases

19 1/21/ The boiling point of Argon is o C. – –Why is it so low? Argon does not interact with other substances because it is so small and has a complete octet of valence electrons. Argon must be made quite cool to allow liquefication via London dispersion forces. – –How does this boiling point prove that London dispersion forces exist? If these forces did not exist, Argon would never liquefy. – –The boiling point of Xenon is o C. Why is it higher than that of Argon? Xenon is bigger and has more electrons than Argon. The likelihood of momentary dipoles is thus greater. It has a greater polarizability than Argon.

20 1/21/ Put the following substances in order from lowest to highest boiling points: – –C 2 H 6 – –NH 3 – –F 2 F 2 can only exhibit intermolecular London forces. C 2 H 6 is not especially polar, but it does have a very slight electronegativity difference between the carbons and the hydrogens. NH 3 exhibits hydrogen bonding, thus giving it a relatively high boiling point. F 2, C 2 H 6, NH 3

21 1/21/ The Liquid State Properties of Liquids

22 1/21/ Surface Tension Surface Tension resistance of liquid to increase in surface area (polar molecules) Increase in attractive forces between molecules at surface compared to forces between molecules in center – –Interior molecules attracted to molecules from every direction No net force on molecule (pulled in all directions) – –Surface molecules dont have molecules to attract it on top side Molecule on surface drawn into bulk of liquid Same total attractive force divided between fewer adjacent molecules, resulting in stronger attractive force of surface molecule toward bulk of liquid Decreases with temperature (reduces intermolecular attractions

23 1/21/ Intermolecular attractive forces act to minimize surface area of liquid – –Geometric shape w/smallest ratio of surface area to volume is sphere – –Surface molecules drawn inward and area of surface of liquid reduced – –Small quantities of liquid form spherical drops – –As drops get larger, weight deforms them into typical tear shape

24 Bubbles (hollow drop)Bubbles (hollow drop) –Surface tension acts to minimize surface (radius of spherical shell of liquid), but opposed by pressure of vapor within bubble –Bubbles in pure water tend to collapse –Bubbles with surfactant are stabilized 1/21/201424

25 1/21/ Surface film – –Smaller surface area causes liquid to behave as though it has skin on surface – –Surface tension enables insects to walk on surface of water – –High intermolecular forces = greater surface tension

26 1/21/ Cohesion-Adhesion Liquids confined within a container have two types of forces present on molecules – –Cohesive forces- intermolecular forces among like molecules – –Adhesive forces-forces between unlike molecules for one another (polar molecules, oxygen in glass attracted to hydrogen in water) Cohesion causes water to form drops, surface tension causes them to be nearly spherical, and adhesion keeps the drops in place

27 Water Adhesive forces greater than cohesive forces MeniscusMeniscus (curved upper surface) is concave (curved downward) Mercury Cohesive forces greater than adhesive forces Meniscus is convex (curved upward) 1/21/201427

28 1/21/ Capillary actionCapillary action – –Instantaneous rising of liquid in narrow tube – –Combination of cohesion/adhesion – –Distance traveled dependent on diameter of tube

29 1/21/ Which would have a higher surface tension, H 2 O or C 6 H 14 ? Why? Would the shape of the H 2 O meniscus in a glass tube be the same or different than C 6 H 14 ? – –Water, having large dipole moment, has relatively large cohesive forces. Hexane is essentially nonpolar so it has low cohesive forces. Water would have higher surface tension. – –Water meniscus is concave because adhesive forces of water to polar constituents on surface of glass are stronger than its cohesive forces. Hexane would have a convex meniscus because it has very small adhesive forces, and the slightly larger cohesive forces would dominate.

30 1/21/ Viscosity Viscosity measure of liquids resistance to flow Increases with intermolecular forces – –To flow, liquids molecules must move past each other – –Move more freely in solutions with relatively low attractive forces – –Nonpolar molecules attracted to each other by only London forces have lower viscosities than polar liquids like water Increases with molecular size Is temperature dependent – –Decrease in viscosity with increased temperature – –Attributed to average molecular KE of liquid which overcomes attractive forces between molecules

31 1/21/ Homework: Read , pp Q pp , #13, 15, 36, 38, 40, 41, 42

32 1/21/ Solids Structure and Types

33 1/21/ Amorphous solids Amorphous solids noncrystalline solids Greek without form – –No orderly structure (arrangement) – –Lack well-defined faces/shapes – –Many are mixtures of particles that don stack together well Plastic, glass, rubber No distinct, sharp melting point, but soften gradually over large temperature range

34 1/21/ Crystalline Solids Atoms, ions, or molecules ordered in well-defined 3-D arrangements (lattice) Unit cellUnit cell: smallest repeating unit of lattice LatticeLattice: unit cells repeated in space in all three dimensions, characteristic of crystalline solid –Lattice point: –Lattice point: part of atom in lattice – –Way spheres arranged in layers determines what type of unit cell we have

35 Primitive or Simple Cube Shape Atom in each corner of cube Atoms in contact along cell edge Stacking Spheres in each layer lay on spheres below/above them –Stacking pattern: AAAAAAA... 1/21/201435

36 Coordination # # atoms/ions surrounding atom/ion in crystal lattice 6 (in red)-has 6 immediate neighbors Value gives measure of how tightly spheres packed together –Larger coordination #, closer spheres to each other –Not close packed (least efficient method-52%) –Very rare packing arrangement for metals (ex. Polonium) Atoms/unit cell 1/21/201436

37 Body Centered Cubic (BCC) Shape Atom at each corner/in center of cube –Atoms in contact along body diagonal Stacking 2 nd layer fit into depressions of 1 st layer/3 rd layer into 2 nd –Stacking pattern: ABABABAB... 1/21/201437

38 Coordination # 8 (each sphere in contact with 4 spheres in layer above/4 below) Not close packed (less efficient packing-68%) Atoms/unit cell 1/21/201438

39 Face-centered Cube (FCC) Shape Atom at each corner/atom in center of each face of cube Stacking Each layer diagonally next to each other. Alternating layer in crevices between spheres. –Stacking pattern: ABABABAB... 1/21/201439

40 Coordination # 12 (more efficient at 74% packing) Not closest packed (Fe, alkali metals) Atoms/unit cell 1/21/201440

41 Determine net number of Na + and Cl - ions in unit cell Na + – –(¼ per edge) x (12 edges) = 3 – –(1 per center) x (1 center) = 1 Cl - – –(1/8 per corner) x (8 corners) = 1 – –(½ per face) x (6 faces) = 3 Correct since # Na = # Cl – –(4 = 4) 1/21/201441

42 42 1/21/2014 Bravais lattice Bravais lattice is innite array of discrete points with arrangement and orientation that appears exactly the same, from whichever points array viewed

43 Introduction to structures and types of solids X-ray analysis of solids 1/21/

44 1/21/ Diffraction Scattering of light from regular array of points or lines Spacing between points/lines/atoms, related to wavelength of light X-rays used because their wavelengths similar to distances between atomic nuclei Reflection of X-rays of wavelength from pair of atoms in two different layers of crystal

45 1/21/ If incident X-ray beam hits crystal lattice, general scattering occurs – destructive interference –Most scattering interferes w/itself and is eliminated (destructive interference) – –(b) Incident rays in phase but reflected rays out of phase: d 2 (difference in distances traveled by two rays) = odd # of half wavelengths

46 1/21/ Diffraction occurs when scattering in certain direction is in phase with scattered rays from other atomic planes – constructive interference –Combine to form waves that reinforce each other (constructive interference) – –(a) Incident/reflected rays in phase: d 1 (difference in distances traveled by two rays ) = whole # of wavelengths

47 X-ray diffraction One of best methods to determine crystal's structureOne of best methods to determine crystal's structure Intense X-ray beam strikes crystalIntense X-ray beam strikes crystal Crystal diffracts X-ray beam differently, depending on structure/orientationCrystal diffracts X-ray beam differently, depending on structure/orientation –Atoms in crystal interact w/x-ray waves to produce interference Diffraction pattern consists of reflections of different intensity used to determine crystals structureDiffraction pattern consists of reflections of different intensity used to determine crystals structure However, many different orientations of crystal collected before true structure of crystal determinedHowever, many different orientations of crystal collected before true structure of crystal determined 1/21/201447

48 481/21/2014

49 49 Bragg Equation Bragg Equation resolution of X-ray diffraction detector Used for analysis of crystal structures – –Each crystalline material has characteristic atomic structure, diffracts X-rays in unique characteristic pattern n = 2d sin d = distance between atoms (unique for each mineral) n = an integer = wavelength of x-rays

50 1/21/ X rays of wavelength 1.54 Å were used to analyze an aluminum crystal. A reflection was produced at Ø = 19.3 degrees. Assuming n = 1, calculate the distance d between the planes of atoms producing this reflection. 2dsinθ = nλ 2(d)(sin 19.3) = 1(1.54 Å) 2(d)(0.3305) = 1.54 Å d = 2.33 Å = 233 pm

51 1/21/ A topaz crystal has a lattice spacing (d) of 1.36 Å (1 Å = 1 x m). Calculate the wavelength of X-ray that should be used if Ø = 15.0 o (assume n = 1) 2dsinθ = nλ 2 (1.36 Å)(.259) = 1(λ) Å = 70.4 pm

52 1/21/ Types of Crystalline Solids (a) Atomic solids (b) Ionic solids (c) Molecular solids

53 1/21/ Structural particles Principal attractive forces between particles Characteristics (physical behavior)Examples AtomsNondirectional covalent bond involving positive ions and mobile valence electrons delocalized throughout crystal (metallic bond) Pure elements All solids at 25 o C except Hg Wide range hardness (electrons move freely from atom to atom, bond metal atoms together with widely varying degrees of force) Wide range melting point (between ionic and covalent compounds) Excellent thermal/electrical conductors (mobile electrons quickly carry charge throughout metal) Malleable/ductile (nondirectional, so stress alters but not destroys crystal) Lustrous (interaction of electrons w/light) K Na Fe Mg Ca Zn Structure and Bonding in Metals

54 1/21/ Closest packing Closest packing most efficient arrangement of spheres Uniform/hard spheres most efficiently use available space Coordination number = 12Coordination number = 12 –Each sphere in contact with 6 spheres in its own layer6 spheres in its own layer 3 spheres in layer above3 spheres in layer above 3 spheres in layer below3 spheres in layer below Stacking – –2 nd layer does not lie directly over those in 1 st layer – –3 rd layer occupies dimples of 2 nd layer in two ways Each sphere in 3 rd layer lies directly over sphere in 1 st layer (aba) Each sphere can occupy positions so that no sphere in 3 rd layer lies over one in 1 st layer (abc)

55 Hexagonal closest packing (hcp) aba arrangement – –Each layer identical to layer two below it – –Ex.: Be, Mg, Ti, Co, Zn, Cd, He (at low T) – –2 atom unit cell (8 x 1/8 + 1) – –Packing efficiency = 74% 1/21/201455

56 Cubic close-packed (ccp) Corresponds to face- centered cube (abc) – –Each layer identical to layer three below it Ex.: Ca, Sr, Ni, Pd, Pt, Cu, Ag, Au, Pb, Pt, Ne/Ar/Kr/Xe (at low T) Packing efficiency = 74% 1/21/201456

57 1/21/ Coordination Numbers for Common Crystal Structures Structure Coordination Number Stacking Pattern simple cubic 6 AAAAAAAA... body-centered cubic 8 ABABABAB... hexagonal closest-packed 12 ABABABAB... cubic closest-packed 12 ABCABCABC.

58 Calculate density of LiF 4.02 Å on edge Same arrangement of ions as NaCl 4(6.94 amu) + 4(19.0 amu) = amu D = amu 1 g (1 Å ) 3 = 2.65 g/cm Å 6.02 x amu (10 -8 ) -3 cm 1/21/201458

59 1/21/ Calculating density of closest packed solid Silver crystallizes in a cubic closest packed structure. The radius of a silver atom is 144 pm. Calculate the desnity of solid silver. Textbook-length of edge of cube: d = r 8 = = 407 pm V = d 3 = (407 pm) 3 = 6.74 x 10 7 pm x 10 7 pm 3 x (1 x cm) 3 /1pm = 6.74 x cm 3 D = m = (4 atoms)(107.9 g/mol)(1 mol/6.022 x atoms) V 6.74 x cm g/cm 3

60 1/21/ The radius of nickel atom is 1.24 Å (1Å = 1 x cm). Nickel crystallizes with a cubic closest packed structure. Calculate the density of solid nickel. d = r 8 = 1.24 Å 8 = 3.51 Å V = d 3 = (3.51 Å) 3 = 43.2 Å Å 3 x (1 x cm) 3 /1pm = 4.32 x cm 3 D = m = (4 atoms)(58.69 g/mol)(1 mol/6.022 x atoms) V 4.32 x cm g/cm 3

61 1/21/ Bonding Models for Metals Electron Sea ModelElectron Sea Model: regular array of metal atoms in sea of electrons Band (Molecular Orbital) ModelBand (Molecular Orbital) Model: electrons assumed to travel around metal crystal in MOs formed from valence atomic orbitals of metal atoms

62 1/21/ Metal Alloys Substance containing mixture of elements/has metallic properties Substitutional alloySubstitutional alloy – –Host metal alloy atoms replaced by other atoms of similar size – –Brass-Cu/Zn, sterling silver-Ag/Cu, pewter-Sn/Cu/Bi/Sb, plumbers solder-Sn/Sb Interstitial alloyInterstitial alloy – –Holes in closest packed metal structure occupied by small atoms – –Steel

63 Network Crystal 1/21/ Structural particlesPrincipal attractive forces between particles Characteristics (physical behavior) Examples Nonmetal atoms (pure elements) Directional covalent bonds leading to giant molecules (metallic bonds) Very hard (brittle) Insoluble in most ordinary liquids Sublime or melt at high temperatures (high MP) Poor thermal and electrical conductors (most insulators) C (diamond, graphite) SiC BN SiO 2 (sand, quartz) DiamondGraphiteQuartz

64 1/21/ Semiconductors Material w/electrical conductivity between conductor and insulator dopingConductivity is enhanced by doping – –In lattice, all atoms bond to 4 neighbors, leaving no free electrons to conduct current (insulator) – –Intentionally introducing impurities into extremely pure semiconductor which allows electrons in bonds to move Make electrons available for conduction (VA) Form holes which can conduct current (IIIA)

65 Polar Molecular Crystals 1/21/ Structural particlesPrincipal attractive forces between particles Characteristics (physical behavior)Examples Discrete polar molecules occupy lattice point Dipole-dipole attractions Low to moderate MP Soluble in some polar liquids Poor thermal/electrical conductors HCl CHCl 3 H 2 S C 6 H 12 O 6 Molecules with H bonded to O, N, F Hydrogen bondsLow to moderate MP Soluble in some hydrogen- bonded and some polar liquids Poor thermal/electrical conductors H 2 O NH 3 HF CH 3 OH

66 1/21/ Nonpolar Molecular Crystals Structural particlesPrincipal attractive forces between particles Characteristics (physical behavior) Examples Atoms (8A), nonpolar molecules Dispersion forcesExtremely low to moderate MP Soluble in nonpolar solvents Poor thermal/electrical conductors Ar Cl 2 H 2 CH 4 I 2 CO 2 CCl 4

67 Ionic Crystals 1/21/ Structural particlesPrincipal attractive forces between particles Characteristics (physical behavior) Examples Positive and negative ions that give strongest attractive forces in chemistry Ion-ion attraction Almost all have rigid lattice Moderate to high MP/BP (large lattice energy required to separate ions) Hard/brittle (when hit, atoms shift position which breaks +/- attraction) Nonconductors as solids but conductors as liquids Many dissolve in water KCl CaF 2 CsBr MgO BaCl 2 NaCl

68 1/21/ Packing arrangement done to minimize anion-anion and cation-cation repulsions Structure of most binary ionic solids explained by closest packing of spheres Anions usually larger than cations packed as hcp or ccp arrangements with cations filling holes Nature of holes depends on anion : cation size – –Trigonal holes formed by 3 spheres in same layer – –Tetrahedral holes formed by sphere sitting in dimple of three spheres in adjacent layer – –Octahedral holes formed between 2 sets of 3 spheres in adjoining layers of closest packed structures Trigonal holes < tetrahedral holes < octahedral holes

69 1/21/ Would AlP have a closest packed structure which is more like NaCl or ZnS? Ionic radii:Would AlP have a closest packed structure which is more like NaCl or ZnS? Ionic radii: –Al 3+ = 50 pm, P 3- = 212 pm –Zn 2+ = 74 pm, S 2- = 184 pm –Na + = 95 pm. Cl - = 181 pm Take anion to cation ratio in each case:Take anion to cation ratio in each case: –S 2- /Zn 2+ = 2.49 (tetrahedral holes) –Cl - /Na + = 1.91 (octahedral holes) –P 3- /Al 3+ = 4.24 (?) Aluminum ions very small compared to phosphorus ion, so not much room is needed. Tetrahedral holes are adequate. So AlP is more like ZnS than NaCl.Aluminum ions very small compared to phosphorus ion, so not much room is needed. Tetrahedral holes are adequate. So AlP is more like ZnS than NaCl.

70 1/21/ Using table 10.7, and based on their properties, classify each of the following substances as to the type of solid it forms.Using table 10.7, and based on their properties, classify each of the following substances as to the type of solid it forms. FeFe Atomic solid with metallic propertiesAtomic solid with metallic properties C 2 H 6C 2 H 6 Contains nonpolar molecules-molecular solidContains nonpolar molecules-molecular solid CaCl 2CaCl 2 Contains Ca 2+ and Cl - ions-ionic solidContains Ca 2+ and Cl - ions-ionic solid GraphiteGraphite Made up of nonpolar carbon atoms covalently bonded in directional planes-network solidMade up of nonpolar carbon atoms covalently bonded in directional planes-network solid F 2F 2 Nonpolar fluorine molecules-molecular solidNonpolar fluorine molecules-molecular solid

71 1/21/ Homework: Read , pp Q pp , #46, 48, 60, 68, 72

72 1/21/ Vapor Pressure And change of state

73 1/21/201473

74 1/21/ Vaporization (Evaporation) Escape of surface liquid molecules to form gas Rate depends on – –Nature of liquid – –Surface area ( SA, evaporation) – –Temperature ( T, proportion of molecules w/ KE above escape energy) Always endothermic (positive) Heat of vaporization (Enthalpy of vaporization, Δ H vap ) -Heat of vaporization (Enthalpy of vaporization, Δ H vap ) -energy required to vaporize one mole of liquid at 1 atm

75 Vapor pressure Develops in gas phase above liquid when liquid placed in closed container When evaporation occurs in closed container, gas molecules cannot escape to surroundings vapor pressureAs more molecules enter gas phase, pressure increases, finally stopping at level (vapor pressure) dependent only on temperature Equilibrium vapor pressure Pressure of vapor present at equilibrium Gas molecules collide w/container walls/liquid –Most re-condense when collide with liquid Initially, evaporation rate greater condensation rate As # gas molecules increase, collisions w/liquid surface increase Rate of evaporation eventually equals to rate of condensation-equilibrium 1/21/201475

76 1/21/ At given temperature, not all molecules moving w/same KE. Small # molecules moving very slow (low KE), while few moving very fast (high KE). Vast majority somewhere in between.

77 1/21/201477

78 1/21/ Variations in Vapor Pressure Related to intermolecular attractive forces – volatile –Liquids w/high intermolecular attraction have relatively low vapor pressures (less volatile-liquids that evaporate readily) – –Liquids w/low intermolecular attraction have relatively high vapor pressures (more volatile) For similar-size molecules – –Hydrogen-bonded substances have largest ΔH vap values (less volatile/lower vapor pressure) – –Polar substances have higher values than similar-size nonpolar substances Vapor pressure increases w/temperature

79 1/21/ Boiling Point Boiling Point vapor pressure of liquid = prevailing atmospheric pressure above that liquid Increasing temperature increases KE increases molecular motion Forces of attraction between molecules (H bonding) disrupted Molecules break free of liquid and become gas At boiling point, liquid turns into gas Normal boiling pointNormal boiling point-BP of liquid at 1 atm

80 1/21/ Clausius-Clapeyron equation Mathematical relationship between heat of vaporization and vapor pressure as measures of intermolecular forces that attract molecules together in liquid state –Relationship between vapor pressure and temperature – –P = vapor pressure – – ΔH vap = heat of vaporization – –R = universal gas law constant – –T = Kelvin temperature – –C = constant (eliminated in 2 nd equation) Based on following assumptions (fail at high P, near critical point) – –Volume of vaporized liquid negligible compared to volume of vapor – –Vapor behaves as ideal gas – – ΔH vao is constant over temperature interval of data – –External pressure doesnt affect vapor pressure

81 1/21/ (a) Vapor pressure as function of temperature Quantitative nature of temperature dependence of vapor pressure Nonlinear (b) Plots of In(P vap ) versus 1/T (Kelvin temperature) Linear

82 1/21/ Clausius-Clapeyron Pg. 487, Sample 10.6 The vapor pressure of water at 25 o C is 23.8 torr, and the heat of vaporization of water is 43.9 kJ/mol. Calculate the vapor pressure of water at 50 o C. ln (23.8 torr) = -43,900 J/mol ( 1 – 1 ) P vapT2 torr J/K mol 323 K 298 K ln (23.8) = P vapT2 torr Take the antilog of both sides to get 23.8 =0.254 = 93.7 torr P vapT2 torr

83 1/21/ Water has a vapor pressure of 24 mmHg at 25 o C and a heat of vaporization of 40.7 kJ/mol. What is the vapor pressure of water at 67 o C? Solution: Simply use the Clausius-Clapeyron Equation to figure out the vapor pressure. We have to be a bit careful about the units of R: the units we're using are kJ, so R = 8.31x10 -3 kJ/mol K. ln(P 2 /P 1 ) = - H vap /R * (1/T 2 - 1/T 1 ) ln(P 2 /24) = kJ/8.31x10 -3 kJ/mol K *(1/340- 1/298) ln(P 2 /24) = 2.03 P 2 /24 = 7.62 P 2 = 182 mmHg

84 1/21/ An unknown liquid has a vapor pressure of 88mmHg at 45 o C and 39 mmHg at 25 o C. What is its heat of vaporization? Solution: Again, use the Clausius- Clapeyron Equation. Here, the only thing we don't know is H vap ln(P 2 /P 1 ) = - H vap /R * (1/T 2 - 1/T 1 ) ln(88/39) = - H vap /8.31x10 -3 *(1/ /298) H vap = 32.0 kJ

85 1/21/ The vapor pressure of 1-propanol at 14.7 o C is 10.0 torr. The heat of vaporization is 47.2 kJ/mol. Calculate the vapor pressure of 1- propanol at 52.8oC. ln(10.0/x) = kJ/mol/ kJ/K mol x (1/326 K – 1/287.9 K) torr

86 1/21/ Changes of State Heating curves Phase diagrams

87 1/21/ T/KE increases and PE constant Only 1 phase present Ice starts to melt at 0 o C T/KE constant due to heat of fusion-80 calories/g (bonds not broken). PE increases until all ice melted Solid/liquid in equilibrium T/KE increases and PE is constant Only 1 phase present Water starts to boil at 100 o C KE constant due to heat of vaporization-539 cal/g PE increases until all water evaporated Liquid/gas in equilibrium T/KE increases, PE is constant Only 1 phase present

88 Heat capacity ( Heat capacity (J/kg- o C -1 ) Amount of energy required to raise temperature of system by 1 o Heat of fusion (Enthalpy of fusion, Δ H fus ) Energy required to convert mole of solid to mole of liquid at constant pressure –Determined as length of first (solid- melting) plateau, which represents heat added, divided by # moles of sample 1/21/ Heat of vaporization (Enthalpy of vaporization, ΔH vap ) Heat absorbed by one mole of liquid when it changes to gas at constant pressure Length of second (liquid- boiling) plateau divided by # moles of sample

89 1/21/ Sublimation Sublimation substance goes directly from solid to gaseous state Reasons for sublimation – –Solids have vapor pressure, but it is normally very low – –Solids with little intermolecular attraction may have substantial vapor pressures and be able to sublime at room conditions. Enthalpy of sublimation (Δh sub ) – –Energy in solid-gas transition – –State function, so ΔH sub = ΔH fus + ΔH vap

90 1/21/ Cooling Curve Reverse-all features same except start with gas and condense to get solid as heat is removed Condensation point/Crystallization point-used in place of BP/MP ΔH cond requires gas to give off heat- always negative (exothermic) ΔH vap = -ΔH cond

91 1/21/ Phase Diagram relationship between pressure and temperature and the three states of matterrelationship between pressure and temperature and the three states of matter

92 1/21/ Fusion curve (solid-liquid line Vapor pressure curve Liquid-gas line Sublimation curve Solid-gas line

93 1/21/ Diagram for closed system Pressure plotted on y-axis/temperature on x-axis Divided into three physical states by three lines meeting at triple point – –Solids-upper left – –Liquids-upper right – –Gases-lower part Along each line is equilibrium mixture of two phases on two sides of that line – –Liquid-solid line usually straight line-changes in pressure have very little effect on solids and liquids (only slightly compressible) – –Gas-solid and liquid-gas lines are curved upward

94 Normal melting pointNormal melting point-temperature at which solid and liquid states have same vapor pressure under conditions where total pressure is 1 atmosphere Normal boiling pointNormal boiling point-temperature at which vapor pressure of liquid is exactly 1 atmosphere SupercoolingSupercooling - process of cooling liquid below its freezing point without its changing to solid SuperheatingSuperheating - process of heating liquid above its boiling point without its boiling 1/21/201494

95 Triple Point Each phase has same temperature and vapor pressure All three phases exist together in equilibrium –Helium-only substance that doesnt have triple point since it has no solid phase Critical Point Maximum temperature at which any liquid can exist, defined by Critical temperature-temp above which substance cant liquefy gas, regardless of how great pressure Critical pressure-pressure above which substance can no longer exist as gas, no matter how high temp –For water: 374°C and 218 atm –Depends on intermolecular attraction (greater intermolecular forces, higher critical temperature) 1/21/201495

96 1/21/ Non-compressible/high density fluid Obtained by either Heating gas above its critical T Compressing liquid at higher P than its critical pressure Above critical point, differences between gases and liquids disappear Density of gas can approach density of liquid phase

97 1/21/ Homework: Read , pp Q pp , #23, 29, 76 (have fun), 78, 88, 90 Do 1 additional exercise and 1 challenge problem Submit quizzes by to me: _zumdahl/ace/launch_ace.html?folder_path=/chemistry/boo k_content/ _zumdahl/ace&layer=act&src=ch10_ ace1.xml _zumdahl/ace/launch_ace.html?folder_path=/chemistry/boo k_content/ _zumdahl/ace&layer=act&src=ch10_ ace1.xml _zumdahl/ace/launch_ace.html?folder_path=/chemistry/boo k_content/ _zumdahl/ace&layer=act&src=ch10_ ace2.xml _zumdahl/ace/launch_ace.html?folder_path=/chemistry/boo k_content/ _zumdahl/ace&layer=act&src=ch10_ ace2.xml _zumdahl/ace/launch_ace.html?folder_path=/chemistry/boo k_content/ _zumdahl/ace&layer=act&src=ch10_ ace3.xml _zumdahl/ace/launch_ace.html?folder_path=/chemistry/boo k_content/ _zumdahl/ace&layer=act&src=ch10_ ace3.xml

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